What should

Given:

Area of the virtual image of each square, A = 6.25 mm²

Area of each square, A₀ = 1 mm²

Hence, the linear magnification of the object can be calculated as:

…(1)

Where, m = magnification

A = Area of virtual image

A₀ = Area of each square

Putting values in equation (1) we get,

m =

⇒ m = 2.5

Also,

m = Image distance(v)/Object distance(u)

m = v/u

Therefore, v = m × u

= 2.5u …(2)

The magnifying glass has a focal length of, f = 10 cm

Applying the lens formula for lens we have:

…(3)

Where, f₀ = focal length of the objective lens

v₀ = Distance of image formation

u₀ = Distance of object from objective

Putting values in the equation (3) we get,

⇒

⇒ u = -(1.5 × 10) / 2.5

Therefore, u = -6 cm

And v = 2.5u

⇒ v = 2.5 × 6

⇒ v = -15 cm

The virtual image cannot be seen by the eyes distinctly, because the image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye.